The equation P=6s represents the perimeter P of a regular hexagon with side length s. What is the
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Hi! You don't actually need to graph this nor find the intersection in order to know the answer. Through the process of elimination we can rule out incorrect answers.
Firstly, we express both equations with y on the left hand side since all choices are in that form:
Knowing these, we can immediately rule out options a, c, and d since they have the wrong equations. This will leave us with option b.
You can verify if the point of intersection provided in option b is true by plugging in the values of x and y to both equations (you will find that it is indeed true).
ANSWER: b. <span>picture of coordinate plane with line y equals negative 2x minus 1 and line y equals negative 3 halves x minus 3. They intersect at 4, negative 9.</span>
Firstly, we express both equations with y on the left hand side since all choices are in that form:
Knowing these, we can immediately rule out options a, c, and d since they have the wrong equations. This will leave us with option b.
You can verify if the point of intersection provided in option b is true by plugging in the values of x and y to both equations (you will find that it is indeed true).
ANSWER: b. <span>picture of coordinate plane with line y equals negative 2x minus 1 and line y equals negative 3 halves x minus 3. They intersect at 4, negative 9.</span>
Given that <span>3 cards are dealt without replacement in a </span><span>standard deck of 52 cards.
Part A:
There are 4 queens in a standard deck of 52 card, thus the probability that the first card is a queen is given by 4 / 52 = 1 / 13.
Since, the first card is not replaced, thus there are 3 queens remaining and 51 ards remaining in total, thus the probability that the second card is a queen is given</span> by 3 / 51 = 1 / 17
Similarly the probability that the third card is a queen is given by 2 / 50 = 1 / 25.
Therefore, the probability that <span>all three cards are queens is given by
Part B:
Yes the probability of drawing a queen of heart is independent of the probability of drawing a queen of diamonds because they are separate cards and drawing one of the cards does not in any way affect the chance of drawing the other card.
Part C:
Given that the first card is a queen, then there are 3 queens remaining out of 51 cards remaining, thus the number of cards that are not queen is 51 - 3 = 48 cards.
Therefore, </span>if the first card is a queen, the probability that the second card will not be a queen is given by 48 / 51 = 16 / 17
Part D:
<span>Given that the first two card are queens, then there are 2 queens remaining out of 50 cards remaining.
Therefore, </span>if two of the three cards are queens ,<span>the probability that you will be dealt three queens</span> is given by 2 / 50 = 1 / 25 = 0.04
Part E:
<span>Given that the first two card are queens, then there are 2 queens remaining out of 50 cards remaining, thus the number of cards that are not queen is 50 - 2 = 48 cards.
Therefore, </span>if two of the three cards are queens ,the probability that the other card is not a queen is given by 48 / 50 = 24 / 25 = 0.96
Part A:
There are 4 queens in a standard deck of 52 card, thus the probability that the first card is a queen is given by 4 / 52 = 1 / 13.
Since, the first card is not replaced, thus there are 3 queens remaining and 51 ards remaining in total, thus the probability that the second card is a queen is given</span> by 3 / 51 = 1 / 17
Similarly the probability that the third card is a queen is given by 2 / 50 = 1 / 25.
Therefore, the probability that <span>all three cards are queens is given by
Part B:
Yes the probability of drawing a queen of heart is independent of the probability of drawing a queen of diamonds because they are separate cards and drawing one of the cards does not in any way affect the chance of drawing the other card.
Part C:
Given that the first card is a queen, then there are 3 queens remaining out of 51 cards remaining, thus the number of cards that are not queen is 51 - 3 = 48 cards.
Therefore, </span>if the first card is a queen, the probability that the second card will not be a queen is given by 48 / 51 = 16 / 17
Part D:
<span>Given that the first two card are queens, then there are 2 queens remaining out of 50 cards remaining.
Therefore, </span>if two of the three cards are queens ,<span>the probability that you will be dealt three queens</span> is given by 2 / 50 = 1 / 25 = 0.04
Part E:
<span>Given that the first two card are queens, then there are 2 queens remaining out of 50 cards remaining, thus the number of cards that are not queen is 50 - 2 = 48 cards.
Therefore, </span>if two of the three cards are queens ,the probability that the other card is not a queen is given by 48 / 50 = 24 / 25 = 0.96